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dimulka [17.4K]

2 years ago

14

A factory produces a batch of toy vehicles, 1/3 of which are cars and the rest are trucks. trucks have defective ratio of 0.02 a

nd cars have defective ratio of 0.01. the quality inspector picks a toy vehicle at random from the batch and finds out that it is defective. what is the probability that the vehicle was a truck

1 answer:

krek1111 [17]2 years ago

4 0

Answer:

Step-by-step explanation:

b

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HELP ME PLEASE......

kirza4 [7]

Answer:

-2

Step-by-step explanation:

intercept at (0, -2)

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B4%7D%20-%20%20%5Cfrac%7B3%7D%7B8%7Dy%20%3D%203%20%5C%5C%20" id="TexFormula

MariettaO [177]

Answer:

(6, - 4 )

Step-by-step explanation:

Given the 2 equations

$\frac{x}{4}$ - $\frac{3}{8}$ y = 3 → (1)

$\frac{5}{3}$ x - $\frac{y}{2}$ = 12 → (2)

Multiply (1) by 8 and (2) by 6 to clear the fractions

2x - 3y = 24 → (3)

10x - 3y = 72 → (4)

Rearrange (3) expressing - 3y in terms of x by subtracting 2x from both sides

- 3y = 24 - 2x

Substitute 3y = 24 - 2x into (4)

10x + 24 - 2x = 72, that is

8x + 24 = 72 ( subtract 24 from both sides )

8x = 48 ( divide both sides by 8 )

x = 6

Substitute x = 6 in either (3) or (4) and solve for y

Substituting in (3)

2(6) - 3y = 24

12 - 3y = 24 ( subtract 12 from both sides )

- 3y = 12 ( divide both sides by - 3 )

y = - 4

Solution is (6, - 4 )

3 0

3 years ago

A produce supplier ships boxes of produce to individual customers. The distribution of weights of shipped boxes is approximately

schepotkina [342]

Answer:

$0.67(4)+ 36$

Option D

Step-by-step explanation:

Given that a producer ships boxes of produce to individual customers.(say x)

X is a normal random variable with mean = 36 and std dev =4 lbs

By definition of std normal variate we know that

$\frac{x-36}{4}$ is N(0,1)

75th percentile of std normal distribution is

0.675

Hence corresponding x would be

$36+0.675(4)$

Option D matches with this value

Hence answer is option D

$0.67(4)+ 36$

5 0

3 years ago

Read 2 more answers

Evaluate the polynomial when a.x=0 and b.x=-1<br> x2 – 3x - 7

Arlecino [84]

Answer:

a) -7 & b) -6

Step-by-step explanation:

Here,

a) x = 0

Solution ↦

$x2 - 3x - 7 \\ = 0 \times 2 - 3 \times 0 - 7 \\ = - 7$

b) x = -1

Solution ↦

$x2 - 3x - 7 \\ = - 1 \times 2 - 3 \times ( - 1) - 7 \\ = - 2 + 3 - 7 \\ = - 9 + 3 \\ = - 6$

6 0

3 years ago

Lori creates the following design for a T-shirt.

kakasveta [241]

<u>Solution-</u>

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7

Area of rectangle 1 = 8.68 sq.in

$\Rightarrow FH \times HI=8.68$

$\Rightarrow EB \times HI=8.68$ (∵ sides of the rectangle 2)

$\Rightarrow 2.8 \times HI=8.68$

$\Rightarrow HI=3.1$

Area of Triangle 1 = 6.48 sq.in

$\Rightarrow \frac{1}{2}\times AE \times EG= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48$ (∵ sides of the rectangle 1)

$\Rightarrow EF+3.1= 5.4$

$\Rightarrow EF=2.3$

$\Rightarrow BH=2.3$ (∵ sides of the rectangle 2)

$BC = BH+HI+IC$

$\Rightarrow 11.7= 2.3+3.1+IC$

$\Rightarrow IC=6.3$

The area of Rectangle 2,

$=EB\times BH =2.8\times 2.3=6.44\ sq.in$

The area of Triangle 2,

$\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in$

The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in

6 0

3 years ago